# Manipulation using Together[]

The use of Together either cancels the common factor from numerator and denominator or multiplies them with a common factor.

In my code I do not want that manipulation.

x[t_] := X0 + X1*E^(I*\[Omega]*t) + Xn1/E^(I*\[Omega]*t)
b[t_] := bb*(E^(I*\[Omega]*t) + E^((-I)*\[Omega]*t))
eq = Numerator[
Together[Derivative[2][x][t] +
c*(Derivative[1][x][t] - Derivative[1][b][t]) + k*(x[t] - b[t]) -
((X1*E^(I*\[Omega]*t) +
Xn1/E^(I*\[Omega]*t))/(\[Alpha]1*X1*
E^(I*\[Omega]*t) + (\[Alpha]n1*Xn1)/E^(I*\[Omega]*t)))*
x[t]]]


This gave me the following output

Manual calculation using

eq1 = Expand[(\[Alpha]1*X1*E^(I*\[Omega]*t) + (\[Alpha]n1*Xn1)/
E^(I*\[Omega]*t))*(Derivative[2][x][t] +
c*(Derivative[1][x][t] - Derivative[1][b][t]) +
k*(x[t] - b[t])) -
(X1*E^(I*\[Omega]*t) + Xn1/E^(I*\[Omega]*t))*x[t]]


produced

Clearly Together has multiplied the numerator and the denominator with a common factor of $$\mathrm{e}^{2\mathrm{i}\omega t}$$.

The following code can be found in this forum. (Is there a version of Together that doesn't cancel common factors?)

Combine[x_] :=
Module[{combinet1, combinet2},
combinet2 =
Together[
x /. Plus -> (If[
FreeQ[{##}, _^_?Negative] && FreeQ[{##}, Rational],
combinet1[##], Plus[##]] &)] /. combinet1 -> Plus]


But this has also failed to produce the intended result.

eq2 = Expand[
Numerator[
Combine[Derivative[2][x][t] +
c*(Derivative[1][x][t] - Derivative[1][b][t]) + k*(x[t] - b[t]) -
((X1*E^(I*\[Omega]*t) +
Xn1/E^(I*\[Omega]*t))/(\[Alpha]1*X1*
E^(I*\[Omega]*t) + (\[Alpha]n1*Xn1)/E^(I*\[Omega]*t)))*
x[t]]]]


This time the common factor is $$\mathrm{e}^{\mathrm{i}\omega t}$$

Any suggestion when the result of $$\frac{a}{b}+\frac{a}{b}$$ can be obtained as $$\frac{ab+ab}{b^2}$$ without any further manipulation so that I can get $$2ab$$ from the numerator?